# Two-Dimensional Elastic Collision of Equal Masses

## Homework Statement

Show that if an elastic collision between a mass and a stationary target of equal mass is not head-on that the projectile and target final velocities are perpendicular. (Hint: Square the conservation of momentum equation, using ##p^2=p\cdot p##, and compare the resulting equation with the energy conservation equation.)

## Homework Equations

##p=mv##
##K=\frac{1}{2}mv^2##

## The Attempt at a Solution

So I followed the hint and got $$\overrightarrow{p_1^2}=\overrightarrow{p_1^{'2}}+\overrightarrow{p_2^{'2}}+2\left(\overrightarrow{p_1}\cdot \overrightarrow{p_2}\right)$$ Plugging in values I got $$m^2\overrightarrow{v_1^2}=m^2\overrightarrow{v_1^{'2}}+m^2\overrightarrow{v_2^{'2}}+2\left(m\overrightarrow{v}_1\cdot m\overrightarrow{v}_2\right)$$ $$\overrightarrow{v_1^2}=\overrightarrow{v_1^{'2}}+\overrightarrow{v_2^{'2}}$$ For the x- and y-components I got the following assuming the target traveled along the x-axis after the collision $$v_1^2cos^2(\theta)=v_1^{'2}cos^2(\theta ^{'})+v_2^{'2}$$ $$v_1^2sin^2(\theta)=v_1^{'2}sin^2(\theta ^{'})$$ This is where I got stuck and wasn't sure how to solve for ##v_1^{'}## or ##v_2^{'}##. I'm not entirely sure when or how to apply the equation for kinetic energy. Am I in the right direction or did I do something wrong?

Last edited:

## Answers and Replies

kuruman
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You did something wrong. The last two equations are dimensionally incorrect. Review their derivation.

You did something wrong. The last two equations are dimensionally incorrect. Review their derivation.
I forgot to square the initial velocity. Were there any other mistakes?

I think I got it.

So the momentum in the x and y-direction is $$v_1=v_1^{'}cos(\theta _1)+v_2^{'}cos(\theta _2)$$ $$0=v_1^{'}sin(\theta _1)-v_2^{'}sin(\theta _2)$$ I then used the following to find the magnitude of the momentum $$p^2=p_x^2+p_y^2$$ Doing this I got $$v_1=v_1^{'2}(cos^2(\theta _1)+sin^2(\theta _1))+v_2^{'2}(cos^2(\theta _2)+sin^2(\theta _2))+2v_1^{'}v_2^{'}(cos(\theta _1)cos(\theta _2)-sin(\theta _1)sin(\theta _2))$$ Using trig identities and the law of cosines, I simplified it to $$v_1^2=v_1^{'2}+v_2^{'2}+2v_1^{'}v_2^{'}cos(\theta _1+\theta _2)$$ Knowing that kinetic energy is conserved in an elastic collision I used the equation for kinetic energy and simplified $$\frac{1}{2}mv_1^2=\frac{1}{2}mv_1^{'2}+\frac{1}{2}mv_2^{'2}$$ $$v_1^2=v_1^{'2}+v_2^{'2}$$ With this formula, ##v_1^2## can be substituted for ##v_1^{'2}+v_2^{'2}## After substituting the equation simplifies to $$0=2v_1^{'}v_2^{'}cos(\theta _1+\theta _2)$$ Solving results in $$v_1^{'}=0$$ $$v_2^{'}=0$$ $$cos(\theta _1+\theta _2)=0$$ Solving for the last equation results in $$\theta _1+\theta _2 = 90°\text{ or }\frac{\pi}{2}$$ Thus, the two velocities must be perpendicular.

haruspex
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Solving results in $$v_1^{'}=0$$ $$v_2^{'}=0$$ $$cos(\theta _1+\theta _2)=0$$
it results in at least one of those three. How do you rule out the first two?

it results in at least one of those three. How do you rule out the first two?
The question asks to prove that the two velocities will be perpendicular, so I assumed the velocities to be non-zero because otherwise there would be no answer as the angle between the two velocities would not exist.

haruspex
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The question asks to prove that the two velocities will be perpendicular, so I assumed the velocities to be non-zero because otherwise there would be no answer as the angle between the two velocities would not exist.
You are given that the collision is not head-on. From that you can show neither velocity is zero.

You are given that the collision is not head-on. From that you can show neither velocity is zero.
That makes sense, thank you very much.

kuruman
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Here is a quick geometric proof, not that your algebraic proof is incorrect.
Momentum conservation for a two equal-mass collision requires that ##\vec{v}_1={\vec{v_1}'}+{\vec{v_2}'}##.
Since the collision is not one-dimensional, if you draw a vector addition diagram, you get a closed triangle with the magnitudes of the three vectors as its sides.
Energy conservation for a two equal-mass collision requires that ##v_1^2= {v'_1}^2+{v'_1}^2##.
Therefore, said triangle obeys the Pythagorean theorem with ##v_1## as the hypotenuse and ##{v'_1}## and ##{v'_2}## as the two right sides.

Here is a quick geometric proof, not that your algebraic proof is incorrect.
Momentum conservation for a two equal-mass collision requires that ##\vec{v}_1={\vec{v_1}'}+{\vec{v_2}'}##.
Since the collision is not one-dimensional, if you draw a vector addition diagram, you get a closed triangle with the magnitudes of the three vectors as its sides.
Energy conservation for a two equal-mass collision requires that ##v_1^2= {v'_1}^2+{v'_1}^2##.
Therefore, said triangle obeys the Pythagorean theorem with ##v_1## as the hypotenuse and ##{v'_1}## and ##{v'_2}## as the two right sides.
Thank you! I wish I had seen that sooner.